What is the impedance of an ideal capacitor with capacitance C at angular frequency ω?
AjωC
B1/(jωC)
Cjω/C
DωC
For a capacitor, i = C·dv/dt. In phasor form, differentiation becomes multiplication by jω, so I = jωC·V, and impedance Z = V/I = 1/(jωC). This can also be written as −j/(ωC), emphasizing that the impedance is purely imaginary and negative — current leads voltage by 90°. Option A (jωC) is the impedance of an inductor, a common mix-up. The key sign difference: inductors have positive imaginary impedance (jωL), capacitors have negative (−j/ωC).
Question 2 Multiple Choice
In AC steady-state analysis using phasors, how does Kirchhoff's Voltage Law (KVL) get applied?
AKVL applies to peak amplitudes only, since all voltages share the same frequency
BKVL applies to phasor voltages (complex numbers), using the same algebraic sum rule as in DC circuits
CKVL requires integration over one complete cycle to handle the phase differences
DKVL applies to RMS values only, with a separate correction for phase
Once voltages are represented as phasors (complex numbers encoding amplitude and phase), KVL says the sum of phasor voltages around any loop equals zero — the exact same algebraic rule as DC. This is the entire point of the phasor transformation: it converts the differential equations of time-domain AC analysis into algebraic equations, so all linear analysis tools (KVL, KCL, nodal analysis, mesh analysis, Thévenin/Norton) apply unchanged, just with complex numbers instead of real ones.
Question 3 True / False
An ideal capacitor connected to an AC source dissipates significant average power because current flows through it continuously.
TTrue
FFalse
Answer: False
A purely reactive element (ideal capacitor or inductor) dissipates zero average power. Average power P = V_rms · I_rms · cos(θ), where θ is the phase angle between voltage and current. For an ideal capacitor, θ = 90° (current leads voltage by 90°), so cos(90°) = 0, giving P = 0. Energy is stored and returned each half-cycle but never dissipated. Only the resistive component of impedance (the real part) dissipates power.
Question 4 True / False
In sinusoidal steady state, every voltage and every current in a linear circuit oscillates at exactly the same frequency as the driving source, though with different amplitudes and phases.
TTrue
FFalse
Answer: True
This is the fundamental property that makes phasor analysis possible. Because all circuit elements are linear (satisfying superposition), a sinusoidal input at frequency ω produces sinusoidal responses at the same frequency ω throughout the circuit. The frequency is preserved; only amplitude and phase change from element to element. If two sources at different frequencies were present, superposition would be applied separately for each frequency.
Question 5 Short Answer
What is the key mathematical insight that allows the differential equations governing capacitors and inductors to be replaced by simple algebraic equations in phasor analysis?
Think about your answer, then reveal below.
Model answer: In phasor analysis, differentiating with respect to time (d/dt) is equivalent to multiplying by jω. This follows from Euler's formula: if v(t) = Re[V·e^(jωt)], then dv/dt = Re[jω·V·e^(jωt)]. So the capacitor equation i = C·dv/dt becomes I = jωC·V in phasor form — a simple algebraic relation. Similarly, the inductor equation v = L·di/dt becomes V = jωL·I. Differentiation becomes multiplication by the constant jω, turning differential equations into linear algebraic equations.
This transformation is the core payoff of phasor analysis. The price is working with complex numbers; the reward is that all differential equations become algebraic, and the entire toolkit of linear circuit analysis applies directly. This is why electrical engineers can solve complicated AC circuits using the same techniques as DC circuits, just with complex-valued impedances instead of real resistances.